Okay, I've finished the multiplication table. Some disturbing features have cropped up. There is a right inverse, but no left inverse and the "group" is not associative.

I've made a mistake somewhere, but I've gone over the calculations in detail and can't find it. Perhaps my definition of the group multiplication is wrong? (That would be far from the first typo I've seen in this text.)

I am trying to construct a group G as the semi-direct product of two groups H ( ) and F ( .) The multiplication tables for each are shown below in the attachment.

In order to do this I need to define the group elements for G. They are simply the direct product (taken in the sense of sets) of the groups F and H:

The multiplication of two elements of G is defined as
where
(I don't know what is standard notation here. Wikipedia would write it as)

First comment: Algebraists tend to write operators on the right, so fT means an element f acted on by a transformation T. Analysts write it the other way round, T(f), thinking of T as a function that acts on the element f. I'm an analyst, so I'll use the Wikipedia notation.

Originally Posted by topsquark

I have chosen as where A and B are two automorphisms on H: as

as

Here's where the trouble starts! An automorphism is a mapping from a group to itself that preserves the group structure. In particular, it must send the identity element to itself. Neither of the maps A, B does this, so these maps are not automorphisms. In fact, H is the cyclic group with four elements; e is the identity element, b has order 2, and the other elements a and c have order 4. There are only two automorphisms of H. One of them is the identity mapping I (that sends each element to itself). The other automorphism, J say, exchanges a and c, and sends e and b to themselves: J(e)=e, J(a)=c, J(b)=b, J(c)=a.

Next, the mapping T is a homomorphism from F to Aut(H). Here again, that means that T has to preserve the group structure. So it has to take the identity element ε of F to the identity element I of the group Aut(H). There are two options for the action of T on x (the other element of F). We could have T(x)=I. But then the action of F on Aut(H) would be trivial, and the semidirect product would just be a direct product. To get a nontrivial semidirect product, we have to take T(x)=J.

If you replace the maps B and A by I and J, and redefine T as above, then you'll find that the semidirect product is indeed a group, with identity element (e,ε). Geometrically, it represents the dihedral group D_8 of symmetries of a square. The subgroup H (or more accurately H×{ε}) is the group of rotations of the square, with F acting on H by conjugating with a reflection.

An automorphism is a mapping from a group to itself that preserves the group structure.

I do seem to remember that comment from somewhere. (It's probably right in front of me in my group theory book and I missed it when I was doing my brief review.) That makes much more sense, especially when I looked over my text's proof that this construction makes an associative group. I had already noted that the text was making some "assumptions" about the homomorphism, but had missed the point about the automorphism.